m∠FDE = 52°
Solution:
Given data:
DE ≅ DF, CD || BE, BC || FD and m∠ABF = 116°
<em>Sum of the adjacent angles in a straight line = 180°</em>
m∠ABF + m∠CBF = 180°
116° + m∠CBF = 180°
m∠CBF = 64°
If CD || BE, then CD || BF.
Hence CD || BE and BE || FD.
Therefore BFCD is a parallelogam.
<em>In parallelogram, Adjacent angles form a linear pair.</em>
m∠CBF + m∠BFD = 180°
64° + m∠BFD = 180°
m∠BFD = 116°
<em>Sum of the adjacent angles in a straight line = 180°</em>
m∠BFD + m∠DFE = 180°
116° + m∠DFE = 180°
m∠DFE = 64°
we know that DE ≅ DF.
<em>In triangle, angles opposite to equal sides are equal.</em>
m∠DFE = m∠DEF
m∠DEF = 64°
<em>sum of all the angles of a triangle = 180°</em>
m∠DFE + m∠DEF + m∠FDE = 180°
64° + 64° + m∠FDE = 180°
m∠FDE = 52°
- I think it's possibly this "The figure must be a square because the diagonals of a square bisect the right angles." but there's also "The figure must be a rhombus because it has exactly 2 pairs of congruent angles." buuut since the first one seems more better i'm assuming then that's probably the right answer.
The measure of angle 1 is 60* since it is 1/6 of 360*.
